# Exercise - Heteroskedasticity-Robust SEs

As we saw in lecture, the command `lm_robust()`

from the `estimatr`

package provides various “flavors” of heteroskedasticity-robust standard errors: HC0, HC1, HC2, and HC3. In this problem you will use simulation to compare the performance of these alternatives in the simplest possible example: a regression model with a single dummy variable \(D_i \in \{0, 1\}\) and normal errors. Suppose that: \[
Y_i = \alpha + \beta D_i + U_i, \quad U_i|D_i \sim \text{indep. N}\big(0, (1 - D_i) \sigma^2 + D_i \big)
\] for \(i = 1, ..., n\). Notice that this design allows for heteroskedasticity. If \(D_i = 1\) then \(U_i\) is standard normal, but if \(D_i = 0\) then \(U_i \sim \text{N}(0, \sigma^2)\). By changing the value of \(\sigma^2\) we can change the extent of heteroskedasticity in the simulation design.

Our goal is to determine which method provides the most accurate approximation to the standard error of \(\widehat{\beta}\), the ordinary least squares estimator of \(\beta\). Because this example is so simple, there’s no need to use `lm_robust()`

or even `lm()`

to calculate \(\widehat{\beta}\). It simply equals the difference of sample means for observations with \(D_i = 1\) compared to those with \(D_i = 0\), i.e. \[
\widehat{\beta} = \bar{Y}_1 - \bar{Y}_0, \quad
\bar{Y_1} \equiv \frac{1}{n_1}\sum_{\{i \colon D_i = 1\}} Y_i, \quad
\bar{Y_0} \equiv \frac{1}{n_0}\sum_{\{i \colon D_i = 0\}} Y_i
\] where \(n\) is the total sample size, \(n_1\) is the number of observations for which \(D_i = 1\) and \(n_0\) is the number of observations for which \(D_i = 0\): \[
n_1 = \sum_{i=1}^n D_i, \quad n_0 = n - n_1.
\] Again, because this example is so simple, we don’t need to use `lm_robust()`

to calculate the standard errors for us. Let \(\text{RSS}_0\) denote the residual sum of squares for the observations with \(D_i=0\) and \(\text{RSS}_1\) denote the total sum of squares for the observations with \(D_i=1\), in other words \[
\text{RSS}_0 \equiv \sum_{\{i \colon D_i = 0\}}^n (Y_i - \bar{Y}_0)^2,\quad
\text{RSS}_1 \equiv \sum_{\{i \colon D_i = 1\}}^n (Y_i - \bar{Y}_1)^2
\] Using this notation, it can be shown that the *usual*, i.e. non-robust, standard error for \(\widehat{\beta}\) is given by \[
\text{SE}_{\text{usual}} = \sqrt{\left(\frac{n}{n_0 n_1}\right) \left( \frac{\text{RSS}_0 + \text{RSS}_1}{n - 2}\right)}.
\] while the various flavors of “robust” standard errors are given by \[
\begin{aligned}
\text{SE}_0 &= \sqrt{\frac{\text{RSS}_0}{n_0^2} + \frac{\text{RSS}_1}{n_1^2}}\\
\text{SE}_1 &= \sqrt{\frac{n}{n-2}\left(\frac{\text{RSS}_0}{n_0^2} + \frac{\text{RSS}_1}{n_1^2}\right)}\\
\text{SE}_2 &= \sqrt{\frac{\text{RSS}_0}{n_0(n_0 - 1)} + \frac{\text{RSS}_1}{n_1(n_1 - 1)}}\\
\text{SE}_3 &= \sqrt{\frac{\text{RSS}_0}{(n_0 - 1)^2} + \frac{\text{RSS}_1}{(n_1 - 1)^2}}
\end{aligned}
\] where \(\text{SE}_0\) corresponds to HC0, \(\text{SE}_1\) to HC1, and so on. **To keep things simple, set \(n = 30\) and \(\alpha = \beta = 0\) in all of your simulations below.** The choice of \(\alpha\) and \(\beta\) doesn’t matter for the results. The choice of \(n\) does matter. I’ve chosen \(n = 30\) so that you can check your results against those from Table 8.1.1 of *Mostly Harmless Econometrics* if you wish. The parameters that will vary in this exercise are \(n_1\) and \(\sigma\).

Write code to calculate each of the standard error estimators from above. Then simulate one draw of the simulation design with \(n_1 = 3\) and \(\sigma = 0.5\) and use it to check your work by comparing against the standard errors provided by

`lm()`

and`lm_robust()`

. They should match*exactly*.Carry out a simulation study with \(n_1 = 3\) and \(\sigma = 0.5\) to answer the following questions:

- What is the mean of the sampling distribution of \(\widehat{\beta}\)?
- What is the
*true*standard error of \(\widehat{\beta}\)? (In other words, what is the standard deviation of the sampling distribution of \(\widehat{\beta}\)?) - Like \(\widehat{\beta}\), each of the five standard errors from above has
*its own*sampling distribution. What are the means and standard deviations of these distributions? - Given a point estimate \(\widehat{\beta}\) and standard error estimate \(\widehat{\text{SE}}\) we can form a t-ratio \(|\widehat{\beta}|/\widehat{\text{SE}}\) to test the null hypothesis \(H_0\colon \beta =0\) against the two-sided alternative. In our simulation design \(\beta = 0\) so a 5% test should reject the null 5% of the time. If you base your test on a standard normal critical value, what is the
*actual*rejection rate of a*nominal*5% test for each of the five standard error estimators?

Repeat the preceding for \(n_1 = 3\) and \(\sigma = 0.85\).

Repeat the preceding for \(n_1 = 3\) and \(\sigma = 1\).

Briefly discuss your results from the preceding three parts.